Test Index

NDA Math Circle

For the next two (2) items that follow :

Consider the two circles

(x-1)^2+(y-3)^2=r^2

and

x^2+y^2-8x+2y+8=0

[NDA I 2016]

Question : 20 of 26

Marks: +1, -0

Solution:

Given equation o f circles

(x-1)^{2}+(y-3)^{2}=r^{2}

(h_{1},k_{1})\equiv

coordinates of centre

\equiv(1,3)

\therefore x^{2}+y^{2}-8x+2y+8=0

\Rightarrow (x-4)^{2}+(y+1)^{2}=(3)^{2}

(h_{2},k_{2})\equiv

coordinates o f centre

\equiv(4,-1)

Radius of circle one

=r_{1}=r

Radius of circle two

=r_{2}=3

∵ Circle intersects at two points so distance between circle is

d < r_{1}+r_{2}

5 < r+3

r > 2

Go to Question:

Prev Question

Next Question